Mathematics · Oct / Nov 2025 · Paper 21 · Question 9
YMA01
As printed in the paper
The question as text
9. In this question you must show detailed reasoning. Solutions relying on calculator technology are not acceptable.
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Show the answer and mark scheme
9(i)[5]
(
)
(
)
2log
4
x
+
5
−
log
x
+
3
=
2
log
=
2
(
)
3
+
3
3
x
3
(
)
2
+
4
x
5
=
9
(
)
x
+
3
2
16
x
+
31
x
−
2
=
0
1
(16
x
−
1)(
x
+
2)
=
0
x
=
only
16
States that log
or else uses rule and proceeds from
a
+
log
b
=
log(
ab
)
9(ii)(a)[3]
given equation log
+
=
+
to log
=
+
a
log
b
log(
a
b
)
ab
log(
a
b
)
b
Deduces ab
−
=
−
=
=
− * M1, A1*
=
a
+
b
ab
a
b
a b
(
1)
b
a
b
1
9(ii)(b)[2]
States either
b > or
1
b ≠ as a would not be defined
1
b > as logs only exist for positive numbers
1
(i)
We are now scoring this question M1, M1, A1, dM1, A1
(
)
(
)
2
M1: Usually scored for the power law of logs
2log
4
x
+
5
→
log
4
x
+
5
(allow without the base
3
3
3)
But may be awarded for writing 2 as
log 9
3
M1: For combining two terms of the original equation.
(
)
2
+
4
x
5
(
)
(
)
E.g
log
=
2
(allow without the base 3) or
2log
4
x
+
5
=
log 9
x
+
3
3
(
)
x
+
3
3
3
A1: A correct equation not involving logs. E.g. (
)
2
+
4
x
5
2
=
or (
)
(
)
9
4
x
+
5
=
9
x
+
3
(
)
+
x
3
dM1: Requires all of the following
2
A starting equation of the form (or equivalent to the form) ()
4
x
+
5
=
k
,
k
(
)
x
+
3
An intermediate equation that is a 3TQA correct attempt to solve the 3TQ by any means condoning the use of a calculatorhere.
It is dependent upon scoring at least one of the previous M marks
A1: CSO 1
16 oe only. The decimal equivalent is 0.0625
___________________________________________________________________________
‘Correct’ solutions with missing or incorrect lines
Example I: Incorrect statement
(
)
2
(
)
2
+
+
log
4
x
5
4
x
5
(
)
(
)
+
−
+
=
3
=
=
leading to full and correct
2log
4
x
5
log
x
3
2
2
9
(
)
(
)
+
+
3
3
log
x
3
x
3
3
solution
Score Special Case M1, M0, A1, dM1, A0 for 3 out of 5
__________________________________________________________________________
Example II: Missing lines
(
)
2
(
)
2
+
+
4
x
5
4
x
5
(
)
(
)
+
−
+
=
=
=
9
2log
4
x
5
log
x
3
2
log
2
(
)
(
)
3
3
3
+
x
+
3
x
3
(
)
2
+
4
x
5
(
)
(
)
Or even
+
−
+
=
=
leading to full and correct solution
2log
4
x
5
log
x
3
2
9
(
)
+
3
3
x
3
can be awarded all the marks.
__________________________________________________________________________
(ii) You can condone the omission of the base
(a) Main method
B1: States or uses log
+
=
o.e. It is implied when log(
=
+
is written down or
a
log
b
log(
ab
)
ab
)
log(
a
b
)
=
+ but it cannot be awarded following incorrect work. Condone log a
for log(
ab
a
b
+
b
if the intention is clear (but the use of this would mean that the A1* mark is not scored)
a
+
b
(
)
Alternatives exist such as
+
−
=
log
a
b
log
b
log(
)
b
M1: Following correct log work, candidate removes the logs, correctly deduces ab
=
a
then attempts to collect terms in a. To award this mark the two terms in a must be moved to
the same side of the equation and the a factorised out. It can be awarded when the term in
(
)
‘a’ is isolated. E.g
±
=
±
=
ab
a
b
a b
1
b
a
+
b
Alternatively removes the logs, deduces
=
and then attempts to collect terms in a
a
b
÷
a
b
b
b
Acceptable alternatives exist. E.g.
=
+
= +
=
−
=
−
ab
a
b
b
1
b
1
a
a
a
b
1
The method mark can be awarded here when there is a single term in a. See double
underlined expression
b
A1*: Correctly proves that
a
=
− via (
a b
−
1)
=
b
or ab
−
a
=
b
.
b
1
This is a given answer and ALL appropriate lines must be seen and be correct.
Condone a missing trailing bracket.
The appropriate lines that must be seen (as a minimum) via the standard approach are;
b
log
ab
=
log(
a
+
b
)
ab
=
a
+
b
a b
(
−
1)
=
b
a
=
−
2
b
b
B1: Substitutes
a
=
−in log
a
+
log
b
and proceeds to
log(
)
b −
b
1
1
b
b
(
)
M1: Substitutes
a
=
−in
log a
+
b
and proceeds to log
+
b
and attempts to write as a
b
1
b
−
1
single fraction
b
A1*: Shows that log
+
=
when
=
−and makes a minimal statement.
a
log
b
log(
ab
)
a
b
1
This is a given answer and all appropriate lines must be seen and be correct (including any
bracketing).
(ii)(b)
B1: For either (i) giving the full restriction on b. E.g.
b >, ‘b is more than one’ or ‘it must be bigger than one’. No (correct) explanation
1
required.
b
or (ii) stating that
b ≠ with a reason such as
1
→∞
or
b −=
1
0
.
b
−
1
Allow reasons such as ‘the denominator cannot be 0’ ‘when b = 1 you get a maths
error’
Do not accept ambiguous statements such as
b ≠because it cannot be 0
1
or (iii) stating that
b … with a reason such as
b −cannot be negative
1
1
B1:
b > or ‘b must be more than one’ as logs only exist for positive numbers.
1
Allow ‘
b >as a must be greater than 0’
1
b
Allow variations that imply this so accept
b > as
1
b −needs to be positive or
b − needs to
1
1
be positive
There really needs to be some words here and the statement and explanation must be given
together and in a logical order. A minimum acceptable response could be ‘as a > 0, b >1’
Do not accept incorrect reasons such as ‘
b > as a must be greater than or equal to 0’
1
Do not award this mark if there are incorrect statements along with correct ones. If unsure
use review.
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